Schwarzschild Radius vs Event Horizon: Black Hole?

[Albertgauss]

TL;DR Summary

Just want to know what the difference between these things is

Real quick, are the terms "Schwarzschild Radius" and "Event Horizon" for a black hole interchangeable or is there some subtle difference between the two? Just looking for a ballpark answer here

 

[PeroK]

Summary:: Just want to know what the difference between these things is

Real quick, are the terms "Schwarzschild Radius" and "Event Horizon" for a black hole interchangeable or is there some subtle difference between the two? Just looking for a ballpark answer here

The event horizon of a non-rotating black hole is a surface at the Schwarzschild Radius:

https://en.wikipedia.org/wiki/Event_horizon#Event_horizon_of_a_black_hole

 

[Dale]

There is an event horizon at the Schwarzschild radius in Schwarzschild coordinates. There are event horizons in other spacetimes besides the Schwarzschild spacetime.

 

[Albertgauss]

Ok, I think I got it. Just to confirm, the event horizon is the surface enclosing the black hole that you can't get out of once you cross, and the radius defining that surface is the Schwarzschild radius. I think the non-rotating, basic, black hole is all I need to consider.

 

[Ibix]

and the radius defining that surface is the Schwarzschild radius.

Why are you asking this? There are a few traps for the unwary here and some context for the question would be useful. For example, you need to be careful with the term "radius" here. It isn't the distance to the center of a black hole - it doesn't have a center.

 

[cianfa72]

The event horizon of a non-rotating black hole is a surface at the Schwarzschild Radius:

In Schwarzschild chart, at a given fixed coordinate time $t$, it is the set of events/points having the same $r=r_s$ and varying $\theta, \phi$ coordinates.

If we let coordinate time $t$ to change then it should be actually a 3D hypersurface.

 

[PeterDonis]

In Schwarzschild chart, at a given fixed coordinate time $t$, it is the set of events/points having the same $r=r_s$ and varying $\theta, \phi$ coordinates.

If we let coordinate time $t$ to change then it should be actually a 3D hypersurface.

Note that Schwarzschild coordinates are singular at $r = r_s$; $t$ is not well-defined there. So you can't "let $t$ change" on the horizon. To see that in spacetime the full event horizon is a 3-surface you need to use some other chart that isn't singular there.

 

[Albertgauss]

I'm actually giving a presentation to explain the difference between "Event Horizon" and "Schwarzschild Radius" to people who do not have any science or math background. So I can't get too technical if I don't have to be. Of course, I understand that may not be possible. The problem is that I've seen these two terms used interchangeably (incorrectly I realize now), but I wanted to know what the difference was.

What I meant by radius was the radius from the singularity out to the event horizon. I'm not sure what's meant by the "black hole doesn't have a center"; I guess I'm confused by that now also; isn't the center a singularity at a point in a space (or region if there are quantum mechanical rules that won't let everything fall exactly to the singularity) that everything that falls into the black hole ends up at? To visualize, I now imagine the Event Horizon as a spherical surface (point of no return) whose surface is defined by the Schwarzschild radius that points from the center (as best as I can say it) of the black hole to the event horizon.

I don't actually know what the following terms are. I'm willing to do a little background reading on them.

"Schwarzschild chart", "If we let coordinate time t to change then it should be actually a 3D hypersurface."

 

[Ibix]

What I meant by radius was the radius from the singularity out to the event horizon.

That's a timelike distance, not a spacelike one. The singularity is more like a point in time than a place in space - once you cross the event horizon, the singularity is your future. In fact, the interior of the black hole can be said to have infinite spatial volume, but a finite extent in time. This is contrary to the naive intuition that it has finite volume but infinite extent in time.

An event horizon is simply the surface dividing regions where light can escape to infinity (future null infinity if you want to get technical) from regions where it cannot. In the case of a Schwarzschild black hole this is a spherical surface of area $4\pi R_S^2$ - but remember that geometry is non-Euclidean, so that does not mean that it contains a spherical volume. In fact, the Schwarzschild radius is related to the maximum time you can survive between event horizon and singularity - about 15$\mu\mathrm{s}$ (edit: and not ms, as I originally wrote) per solar mass of black hole, from memory.

 

[PeterDonis]

about 15ms per solar mass of black hole

It's actually about 7 microseconds. The simplest heuristic estimate for maximum time from horizon to singularity is

$$
\tau = \frac{2}{3} ( 2M )
$$

in units where $G = c = 1$. For the Sun, $2M$ is about 3 kilometers, which is about 10 microseconds of time (just divide by $c$), and 2/3 of that is about 7 microseconds.

 

[Albertgauss]

I think Ibix reply is going to work for what I need. I think people with non-science, at the level of high school math or less could understand that. The correct notion of a "the singularity is more like a point in time than a place in space - once you cross the event horizon, the singularity is your future"--even sounds cooler than the spatial volume I started this thread with. Of course, if anyone else has anything to add, that's always good, but I'm going to work with Ibix reply, which is, of course, the culmination of everyone's efforts here I really appreciate.

 

[berkeman]

I think the non-rotating, basic, black hole is all I need to consider.

I think if I were in your audience at that presentation, it would be natural for me to ask "What about for a rotating black hole?"

https://en.wikipedia.org/wiki/Rotating_black_hole

 

[Ibix]

It's actually about 7 microseconds. The simplest heuristic estimate for maximum time from horizon to singularity is

$$
\tau = \frac{2}{3} ( 2M )
$$

in units where $G = c = 1$. For the Sun, $2M$ is about 3 kilometers, which is about 10 microseconds of time (just divide by $c$), and 2/3 of that is about 7 microseconds.

This paper provides a detailed calculation - the result is in equation 18. Note that, confusingly, they use $r_S$ for the radius at which the object was at rest. In the limit as $r_S\rightarrow 2m$ the answer is $\tau\rightarrow \pi m$. That works out to about 15 microseconds per solar mass. Your figure is for infall from rest at infinity, I think.

But yes, I was off by three orders of magnitude - will correct my post above.

 

[PeterDonis]

Your figure is for infall from rest at infinity, I think.

Yes. Whereas the paper you reference is for free fall from rest at a finite radius, and taking the limit of that as the finite radius approaches $2M$. I agree the latter is a better number if you are not just making a quick heuristic estimate.

 

[PeroK]

It's actually about 7 microseconds. The simplest heuristic estimate for maximum time from horizon to singularity is

$$
\tau = \frac{2}{3} ( 2M )
$$

in units where $G = c = 1$. For the Sun, $2M$ is about 3 kilometers, which is about 10 microseconds of time (just divide by $c$), and 2/3 of that is about 7 microseconds.

I think that's the time to fall from the event horizon to the singularity if you start from infinity. The time to fall from the event horizon, starting from rest at the horizon is$$\tau = \pi M \approx 16\mu s$$

 

[cianfa72]

In the case of a Schwarzschild black hole this is a spherical surface of area $4\pi R_S^2$ - but remember that geometry is non-Euclidean, so that does not mean that it contains a spherical volume.

Can you elaborate it, please ? AFAIK event horizon is a 2D spherical surface 'labeled' by fixed coordinate $t$ and $r=r_s$ and varying $\theta, \phi$. At fixed $t$ we get from Schwarzschild metric a 3D spacelike hypersurface that can be foliated by 2D spherical surfaces (event horizon is just one of these 2D spherical surfaces).

If the above is correct, the geometry of the spacelike hypersurface at fixed $t$ is non-Euclidean (Flamm geometry): for this reason you said it does not contain a spherical volume ?

In fact, the Schwarzschild radius is related to the maximum time you can survive between event horizon and singularity - about 15$\mu\mathrm{s}$ (edit: and not ms, as I originally wrote) per solar mass of black hole, from memory.

Is actually this calculated time the proper time elapsed on your wristwatch between crossing event horizon and singularity ?

 

[PeterDonis]

I think that's the time to fall from the event horizon to the singularity if you start from infinity. The time to fall from the event horizon, starting from rest at the horizon is$$\tau = \pi M \approx 16\mu s$$

Yes. See posts #13 and #14.

 

[Ibix]

Interesting that it's only a factor of ~2 difference between freefall from rest at infinity and from rest as near to "on the horizon" as you can get and still be at rest.

 

[PeterDonis]

AFAIK event horizon is a 2D spherical surface 'labeled' by fixed coordinate $t$ and $r=r_s$ and varying $\theta, \phi$.

No. Schwarzschild coordinates are singular at the horizon. $t$ is not well-defined there.

At fixed $t$ we get from Schwarzschild metric a 3D spacelike hypersurface that can be foliated by 2D spherical surfaces (event horizon is just one of these 2D spherical surfaces).

No. Schwarzschild coordinates are singular at the horizon. The spacelike surfaces of fixed $t$ that you describe here only cover the region outside the horizon. Here we are talking about inside the horizon.

You can adopt Schwarzschild coordinates inside the horizon (using a coordinate patch that is completely disconnected from the Schwarzschild coordinate patch outside the horizon), but surfaces of fixed $t$ inside the horizon are not spacelike 3-surfaces. They are surfaces with two spacelike and one timelike tangent vector, which can be thought of as a timelike sequence of spacelike 2-spheres.

 

[PeterDonis]

Is actually this calculated time the proper time elapsed on your wristwatch between crossing event horizon and singularity ?

The times described in post #13 and #14 are proper times elapsed on a clock between crossing the horizon and reaching the singularity, for two cases: free fall from rest at infinity, and free fall from rest an infinitesimal distance above the horizon.

 

[Ibix]

Can you elaborate it, please ? AFAIK event horizon is a 2D spherical surface 'labeled' by fixed coordinate $t$ and $r=r_s$ and varying $\theta, \phi$. At fixed $t$ we get from Schwarzschild metric a 3D spacelike hypersurface that can be foliated by 2D spherical surfaces (event horizon is just one of these 2D spherical surfaces).

The event horizon is labelled by $r=r_S=2GM/c^2$, yes, but the Schwarzschild $t$ coordinate is not defined on the horizon, so you can't talk about $t$ there. The best representation of the Schwarzschild spacetime is the Kruskal diagram, which is worth a look - it's in Carroll's lecture notes, for example. It shows how the $t$ coordinate fails and how the $r$ coordinate behaves both inside and outside the horizon.

If the above is correct, the geometry of the spacelike hypersurface at fixed $t$ is non-Euclidean (Flamm geometry): for this reason you said it does not contain a spherical volume ?

Note that the Flamm surface isn't defined on or in $r=r_S$ - that's another example of the problem here.

Is actually this calculated time the proper time elapsed on your wristwatch between crossing event horizon and singularity ?

Yes. As Peter notes, the two different times (7 and 15$\mu\mathrm{s}$) are for different initial conditions. You can also index concentric spheres inside the horizon the same way as you do outside, and this is a time-like indexing. But just as $\Delta r$ isn't really a distance anybody would measure outside, it isn't a time anyone would measure inside.

 

[cianfa72]

The event horizon is labelled by $r=r_S=2GM/c^2$, yes, but the Schwarzschild $t$ coordinate is not defined on the horizon, so you can't talk about $t$ there.

As far as I can understand such $t$ coordinate singularity at the horizon is like the spherical polar coordinates on a 2-sphere at North Pole. At this point one of the two coordinates is not defined since it maps to a continuous infinite 'range of values' of such coordinate.

In the same way coordinate $t$ is not defined at event horizon (i.e. on events/points spatially located at it).

 

[PeterDonis]

As far as I can understand such $t$ coordinate singularity at the horizon is like the spherical polar coordinates on a 2-sphere at North Pole.

No, it's not. The best way I know of to see what is actually happening with Schwarzschild coordinates at the horizon is to look at how a Schwarzschild coordinate "grid" looks in Kruskal coordinates. I discuss this in this Insights article:

https://www.physicsforums.com/insights/schwarzschild-geometry-part-3/

Notice, in particular, where all of the lines of constant Schwarzschild coordinate $t$ intersect. Except for that one point (which isn't even present in a realistic model where a black hole forms by gravitational collapse), there are no points on the horizon that any line of constant $t$ passes through.

In the same way coordinate $t$ is not defined at event horizon (i.e. on events/points spatially located at it).

The horizon is a null surface, meaning it moves outward at the speed of light, so it makes no sense to even talk about events "spatially located" at it. It's not a place in space.

 

[Ibix]

As far as I can understand such $t$ coordinate singularity at the horizon is like the spherical polar coordinates on a 2-sphere at North Pole. At this point one of the two coordinates is not defined since it maps to a infinite continuous 'range of values' of such coordinate.

In the same way coordinate $t$ is not defined at event horizon (i.e. on events/points spatially located at it).

It's not exactly the same. The pole of a sphere is only one point, while the event horizon is a null surface. Even if you fix $\theta$ and $\phi$, it's a null line. So on a sphere you can mildly abuse latitude and longitude and talk about 90°N where the longitude doesn't matter because all longitudes are the same point, and only mathematicians and other pedants will get upset. But you can't do the same in Schwarzschild spacetime with $r=r_S$ because you do need another coordinate in the surface of the event horizon to distinguish the events that make it up - and $t$ doesn't work there.

 

[Ibix]

The best tool for understanding a Schwarzschild black hole is the Kruskal diagram. Here's one copied from Wikipedia's article on Kruskal-Szekeres coordinates (it's CC licensed requiring attribution - so thanks to @DrGreg):

Kruskal-Szekeres coordinates have the neat property that radially moving light follows 45° lines, just like on a Minkowski diagram. This means that timelike directions are always steeper than that.

The red region labelled I is the exterior. Someone hovering at constant $r,\theta,\phi$ follows the red hyperbolae on this diagram, crossing the straight lines which are planes of constant $t$. The event horizon is the boundary between the red and blue regions. It is an outgoing null surface as you can tell from its gradient, and you can see that $t$ is ill-defined there. The interior of the hole is the blue region labelled II, and here you can see that the lines of constant $r$ are now spacelike, while lines of constant $t$ are timelike. So because you always travel in a timelike direction you are forced to ever smaller $r$ until you hit the singularity. In principle, though, you are free to move as far as you like in the $t$ direction, which is why I described the interior as infinite - those hyperbolae extend to infinity.

Regions III and IV don't exist in a realistic black hole - they are artifacts of the idealised Schwarzschild "eternal black hole, came from nowhere, never changes" model. III is another exterior region, disconnected from region I, and IV is a white hole, which always lies in the past wherever you are. Anything in that interior region is forced to move into one of the exteriors, or directly into the black hole through the origin of the diagram. As I say, these are essentially the mathematics' way of saying "an eternal black hole with no origin? Pull the other one mate" and don't appear in more realistic models.

 

[ergospherical]

Can you elaborate it, please ? AFAIK event horizon is a 2D spherical surface 'labeled' by fixed coordinate $t$ and $r=r_s$ and varying $\theta, \phi$. At fixed $t$ we get from Schwarzschild metric a 3D spacelike hypersurface that can be foliated by 2D spherical surfaces (event horizon is just one of these 2D spherical surfaces)

Sort of, but the event horizon of a black hole $B$ is the entire 3d boundary $\dot{B}$ and not just its intersection with a suitable spacelike hypersurface and in these sorts of problems they pretty much always get you to use some sort of advanced time coordinate parameterising the null generators along with a couple of spatial coordinates $y^{\mu}$ on the 2d subspaces transverse to the generators.

 

[PeroK]

This paper provides a detailed calculation - the result is in equation 18. Note that, confusingly, they use $r_S$ for the radius at which the object was at rest. In the limit as $r_S\rightarrow 2m$ the answer is $\tau\rightarrow \pi m$. That works out to about 15 microseconds per solar mass. Your figure is for infall from rest at infinity, I think.

But yes, I was off by three orders of magnitude - will correct my post above.

There was a previous thread that included some of these calculations:

https://www.physicsforums.com/threads/how-long-does-it-take-to-fall-into-a-black-hole.1000235/

To add a bit more:

1) If we start from rest at infinity on a radial plunge, then the proper time to fall from $r_1$ to $r_2$ is:
$$\Delta \tau = \int_{r_2}^{r_1} \sqrt{\frac r {2M}} dr = \frac{2}{3\sqrt{2M}}\big (r_1^{3/2} - r_2^{3/2} \big )$$If we let $r_2 = 0$, then:$$\Delta \tau_0 = \frac{2r_1}{3}\sqrt{\frac{r_1}{2M}}$$And, if $r_1 = 2M$, then:$$\Delta \tau_0 = \frac{4M}{3}$$
2) If we start from rest at $r_1$, then the proper time to fall to $r_2$ is:
$$\Delta \tau = \int_{r_2}^{r_1} \frac{dr}{\sqrt{\frac{2M}{r} - \frac{2M}{r_1}}} = \frac{1}{\sqrt{2M}}\int_{r_2}^{r_1} \frac{dr}{\sqrt{\frac{1}{r} - \frac{1}{r_1}}}$$The integral can be done with the substitution $u = \sqrt{\dfrac{1}{r} - \dfrac{1}{r_1}}$ which yields:
$$\Delta \tau = \frac{r_1}{\sqrt{2M}}\bigg (\sqrt{r_2}\sqrt{1 - \frac{r_2}{r_1}} + \sqrt{r_1}\tan^{-1}\big ( \sqrt{\frac{r_1}{r_2} -1} \big ) \bigg )$$For $r_2 = 0$ we get:
$$\Delta \tau_0 = \frac{\pi r_1}{2}\sqrt{\frac{r_1}{2M}}$$And, if $r_1 = 2M$:
$$\Delta \tau_0 = \pi M$$

 

[cianfa72]

The red region labelled I is the exterior. Someone hovering at constant $r,\theta,\phi$ follows the red hyperbolae on this diagram, crossing the straight lines which are planes of constant $t$. The event horizon is the boundary between the red and blue regions. It is an outgoing null surface as you can tell from its gradient, and you can see that $t$ is ill-defined there.

ok, points (events) on the event horizon (as boundary between red and blue regions) have well-defined $T,X$ Kruskal coordinates constrained by $T=X$ and varying $\theta, \phi$. Therefore it is a 3-surface (there are three 'free' coordinates to map its points). It is a null 3-surface since tangent vectors at each point are lightlike (null).

Said that I've however some difficulty to grasp the physical meaning of event horizon. At fixed Kruskal coordinate time $T_0$ it is 2-surface 'labeled' by $X=T=T_0$. As 'subset' of a null 3-surface is it a null 2-surface itself ?

 

[PeterDonis]

As 'subset' of a null 3-surface is it a null 2-surface itself ?

No. It's a spacelike 2-surface. Only one of the "dimensions" of the horizon 3-surface is null, not all three.

 

[cianfa72]

No. It's a spacelike 2-surface. Only one of the "dimensions" of the horizon 3-surface is null, not all three.

Ah ok, so as spacelike 2-surface it actually represents '2-sphere space' at Kruskal coordinate time $T_0$ and $X=T_0$ (i.e. the subset of events simultaneous in KS coordinate chart having $X=T=T_0$), I believe.

 

[PeterDonis]

as spacelike 2-surface it actually represents '2-sphere space' at Kruskal coordinate time $T_0$ and $X=T_0$ (i.e. the subset of events simultaneous in KS coordinate chart having $X=T=T_0$), I believe.

Yes. Every point on the Kruskal chart represents a spacelike 2-sphere.